7
$\begingroup$

In the paper On the Castelnuovo-Mumford regularity of the cohomology ring of a group, Symonds describes the following space.

Let $G = (\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,ab\}$ be an elementary abelian $2$-group, and let $Z = \{ z_1, \dots, z_6\}$ be a discrete space with six elements. Let $G$ act on $Z$ such that each rank one subgroup of $G$ is a stabilizer of some element of $Z$. If I'm not mistaken, this means that the action is for example given as follows: $$a \cdot z_1 = b \cdot z_1 = z_2;\\ a \cdot z_3 = z_4; \; b \cdot z_3 = z_3; \; b \cdot z_4 = z_4;\\ b \cdot z_5 = z_6; \; a \cdot z_5 = z_5; \; a \cdot z_6 = z_6.$$

Let $SZ$ be the suspension of $Z$, with the induced $G$-action.

Question. How to compute the equivariant cohomology $H^*_G(SZ)$ as an $H^*_G$-module (with coefficients $\mathbb{F}_2$)?

I believe I could compute the equivariant cohomology $H^*_G(Z)$, using the Mayer–Vietoris sequence for equivariant cohomology (it split as a disjoint union of three $G$-spaces), the Künneth theorem, and the known results about the equivariant cohomology of spaces with a free (resp. trivial) action. But unfortunately there's no suspension isomorphism for equivariant cohomology, as far as I can tell...

Of course I know the definition of equivariant cohomology; the classifying space $BG$ is $(\mathbb{RP}^\infty)^2$, and $H^*_G = \mathbb{F}_2[x,y]$ is a polynomial algebra on two variables of degree $1$. The total space of the universal bundle is $(S^\infty)^2$ with $a$ (resp. $b$) acting by the antipodal action on the first (resp. second) factor, and $H^*_G(SZ) = H^*(EG \times_G SZ)$. But this isn't exactly helpful, or at least I don't see how to compute the equivariant cohomology just from that.

$\endgroup$
  • $\begingroup$ Maybe it's better ask on MathOverflow. $\endgroup$ – Vincenzo Zaccaro Apr 11 '17 at 9:48
6
$\begingroup$

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form $$ E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2). $$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable.

There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(SZ;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$

However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module.

I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.

$\endgroup$
  • $\begingroup$ Thanks! As a neophyte in equivariant cohomology, your answer is very helpful. I'll try to work out the details and see if I haven't missed anything. $\endgroup$ – Najib Idrissi Apr 13 '17 at 14:13
  • $\begingroup$ @NajibIdrissi: You're welcome. I should add that I've had a change of mind regarding the last sentence of my answer. It would be true (that $H^*_G(SZ;\mathbb{F}_2)$ is free as a $H^*_G$-module) if $G$ acts trivially on $H^1(SZ;\mathbb{F}_2)$. $\endgroup$ – Mark Grant Apr 16 '17 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.